Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

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The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const int N = 100010;
int ara[104][104];
int n, m;
vector<int> ans[3];
int check(int k) {
  int cost = k;
  for (int i = 0; i < m; i++) cost += ara[0][i] - k;
  for (int i = 1; i < n; i++) {
    int first = ara[i][0] - (ara[0][0] - k);
    for (int j = 0; j < m; j++) {
      int c = ara[i][j] - (ara[0][j] - k);
      if (c != first || c < 0) return 1000000000;
    }
    cost += first;
  }
  return cost;
}
void print(int k) {
  for (int i = 0; i < k; i++) cout << "row " << 1 << endl;
  for (int i = 0; i < m; i++)
    for (int j = 0; j < ara[0][i] - k; j++) cout << "col " << i + 1 << endl;
  for (int i = 1; i < n; i++) {
    for (int j = 0; j < ara[i][0] - (ara[0][0] - k); j++)
      cout << "row " << i + 1 << endl;
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < m; j++) cin >> ara[i][j];
  int mn = *min_element(ara[0], ara[0] + m);
  int ans = -1;
  int ans_c = 1000000000;
  for (int k = 0; k <= mn; k++) {
    int cost = check(k);
    if (cost < ans_c) {
      ans_c = cost;
      ans = k;
    }
  }
  if (ans == -1)
    cout << ans << endl;
  else {
    cout << ans_c << endl;
    print(ans);
  }
}